Lie Groups and Lie Algebras

InteractiveHall Ch. 2 §§1–7

Matrix Lie groups

The groups of continuous symmetry that matter in practice (rotations, unitary matrices, the Lorentz group) are groups of matrices, closed under limits. Treating them as subsets of $\mathbb C^{n^2}$ lets us ask topological questions first: is the group compact, connected, simply connected? Those three answers control most of what follows.

The idea

Matrix Lie group (Hall Ch. 2 §1). A subgroup

G\subseteq\mathrm{GL}(n,\mathbb C)

that is closed in

\mathrm{GL}(n,\mathbb C)

: if

A_m\in G

and

A_m\to A

with

A

invertible, then

A\in G

. Most groups are cut out by polynomial equations, which are automatically closed conditions.

The standard list (Hall Ch. 2 §2).

\mathrm{GL}(n)

;

\mathrm{SL}(n)

(

\det=1

);

\mathrm O(n)

(

A^\top A=I

) and

\mathrm{SO}(n)

;

\mathrm U(n)

(

A^*A=I

) and

\mathrm{SU}(n)

; the Lorentz groups

\mathrm O(n,1)

preserving

x_1^2+\dots+x_n^2-t^2

; the symplectic groups; the Heisenberg group of upper triangular

3\times3

matrices with 1s on the diagonal; the Euclidean groups

\mathrm E(n)

of maps

x\mapsto Rx+b

, written as

\begin{pmatrix}R&b\\0&1\end{pmatrix}

.

Compact, connected, simply connected (Hall Ch. 2 §§3–5). A matrix group is compact when it is closed and bounded in

M_n(\mathbb C)

:

\mathrm O(n),\mathrm{SO}(n),\mathrm U(n),\mathrm{SU}(n)

are,

\mathrm{GL}(n)

,

\mathrm{SL}(n,\mathbb R)

and the Lorentz group are not.

\mathrm{SO}(n)

,

\mathrm U(n)

,

\mathrm{SU}(n)

,

\mathrm{SL}(n)

and

\mathrm{GL}(n,\mathbb C)

are connected;

\mathrm{GL}(n,\mathbb R)

and

\mathrm O(n)

have two components, told apart by the sign of the determinant.

\mathrm{SU}(2)

is simply connected, while

\mathrm{SO}(3)

and the circle are not.

Lie groups in general (Hall Ch. 2 §7). A Lie group is a smooth manifold with a group structure whose multiplication and inversion are smooth. Every matrix Lie group is one (a closed subgroup of a Lie group is an embedded submanifold, by Cartan's theorem), so the tools of the Smooth Manifolds page apply: tangent spaces, vector fields, flows. Not every Lie group is a matrix group; the universal cover of

\mathrm{SL}(2,\mathbb R)

is the standard counterexample.

Explore: which groups contain this matrix?

Drag the tips of the two columns

Ae_1

and

Ae_2

. The shaded parallelogram is the image of the unit square; the dashed circle is the unit circle.

Snap to an element of

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Matrix Lie group: A group of invertible matrices that contains every invertible limit of its own elements.
Orthogonal group: The real matrices that preserve lengths and angles: rotations together with reflections.
Unitary group: The complex matrices that preserve the standard Hermitian inner product, the complex counterpart of rotations and reflections.
Special: The prefix that means "determinant 1": volume-preserving and, for real matrices, orientation-preserving.
Compact: Closed and bounded, for a group of matrices: no sequence of elements can run off to infinity or escape to a limit outside the group.
Connected component: A maximal piece of the group within which any two elements can be joined by a continuous path of elements.
Simply connected: Describes a connected space in which every loop can be shrunk to a point without leaving the space.
Heisenberg group: Upper triangular 3 by 3 matrices with ones on the diagonal: the simplest group in which the order of multiplication matters only through one central number.

Examples

Example 1.1 (the circle).

$\mathrm{SO}(2)=\Big\{\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\Big\}\cong\mathrm U(1)=\{e^{i\theta}\}$ : compact, connected, abelian, and not simply connected (the loop $\theta\in[0,2\pi]$ does not shrink).

Example 1.2 (

\mathrm{SU}(2)

is a 3-sphere).

$\mathrm{SU}(2)=\Big\{\begin{pmatrix}a&-\bar b\\b&\bar a\end{pmatrix}:|a|^2+|b|^2=1\Big\}$ (Exercise 1.1). Writing $a,b$ in real coordinates identifies it with the unit sphere $S^3\subset\mathbb R^4$ , which is simply connected.

Example 1.3 (two components).

$\det:\mathrm{GL}(n,\mathbb R)\to\mathbb R\setminus\{0\}$ is continuous and onto a disconnected set, so $\mathrm{GL}(n,\mathbb R)$ is disconnected. The matrices with $\det>0$ form the identity component $\mathrm{GL}^+(n,\mathbb R)$ (Exercise 1.3). In the figure, no drag can carry the parallelogram from one orientation to the other without collapsing it.

Example 1.4 (a non-compact group).

$\mathrm{diag}(t,1/t)\in\mathrm{SL}(2,\mathbb R)$ for every $t>0$ , and its norm is unbounded, so $\mathrm{SL}(2,\mathbb R)$ is not compact. Lorentz boosts $\begin{pmatrix}\cosh s&\sinh s\\\sinh s&\cosh s\end{pmatrix}$ do the same in $\mathrm O(1,1)$ .

Example 1.5 (a subgroup that is not closed).

For irrational $a$ , $\{\mathrm{diag}(e^{it},e^{iat}):t\in\mathbb R\}$ is a subgroup of the torus $\mathrm U(1)\times\mathrm U(1)$ that winds around it densely. Its closure is the whole torus, so it is not a matrix Lie group (Hall Ch. 2 §1). Such "dense windings" are why the definition asks for closedness.

Example 1.6 (the Euclidean group).

$\begin{pmatrix}R&b\\0&1\end{pmatrix}\begin{pmatrix}x\\1\end{pmatrix}=\begin{pmatrix}Rx+b\\1\end{pmatrix}$ , so rigid motions of $\mathbb R^n$ are a matrix group, $\mathrm E(n)\cong\mathbb R^n\rtimes\mathrm O(n)$ . Multiplying two of these matrices composes the motions.

Example 1.7 (the Heisenberg group).

$\begin{pmatrix}1&a&c\\0&1&b\\0&0&1\end{pmatrix}\begin{pmatrix}1&a'&c'\\0&1&b'\\0&0&1\end{pmatrix}=\begin{pmatrix}1&a+a'&c+c'+ab'\\0&1&b+b'\\0&0&1\end{pmatrix}$ . It is diffeomorphic to $\mathbb R^3$ , and non-commutative only through the corner entry $ab'$ versus $a'b$ .

Check your understanding

Is $\mathrm{GL}(2,\mathbb R)$ connected?

Yes: it is an open subset of $\mathbb R^4$ .Open subsets can be disconnected; this one is.
No: the determinant is a continuous map onto $\mathbb R\setminus\{0\}$ .

A path from

\det>0

to

\det<0

would have to pass through

\det=0

, which is not in the group.

Which of these groups is compact?

$\mathrm{SL}(2,\mathbb R)$ Example 1.4.
$\mathrm{SO}(3)$
The Heisenberg groupIt is a copy of $\mathbb R^3$ .

\mathrm{SO}(3)

is closed and every column of an orthogonal matrix has length 1, so it is bounded.

Which statement is true?

$\mathrm{SU}(2)$ is simply connected and $\mathrm{SO}(3)$ is not.
Both are simply connected.The full turn in $\mathrm{SO}(3)$ is a loop that does not shrink (module 6).
Neither is. $\mathrm{SU}(2)$ is a 3-sphere.

Module 6 shows that

\mathrm{SU}(2)\to\mathrm{SO}(3)

is a double cover, which is how one sees

\pi_1(\mathrm{SO}(3))=\mathbb Z/2

.

Exercises

Exercise 1.1The shape of

\mathrm{SU}(2)

Show that every element of $\mathrm{SU}(2)$ has the form $\begin{pmatrix}a&-\bar b\\b&\bar a\end{pmatrix}$ with $|a|^2+|b|^2=1$ , and conversely.

Hint

For $U=\begin{pmatrix}a&c\\b&d\end{pmatrix}$ , $U^{-1}=U^*$ and $\det U=1$ give two expressions for $U^{-1}$ .

Solution

With $\det U=1$ , $U^{-1}=\begin{pmatrix}d&-c\\-b&a\end{pmatrix}$ , and unitarity says this equals $U^*=\begin{pmatrix}\bar a&\bar b\\\bar c&\bar d\end{pmatrix}$ . So $d=\bar a$ , $c=-\bar b$ , and $\det U=|a|^2+|b|^2=1$ . Conversely, such a matrix has orthonormal columns and determinant 1.

Exercise 1.2

\mathrm O(n)

and

\mathrm U(n)

are compact

Show that $\mathrm O(n)$ and $\mathrm U(n)$ are compact.

Hint

Closed: $A\mapsto A^*A$ is continuous. Bounded: every column is a unit vector.

Solution

$\mathrm U(n)$ is the preimage of $\{I\}$ under the continuous map $A\mapsto A^*A$ , hence closed in $M_n(\mathbb C)$ . Its columns are orthonormal, so every entry has modulus at most 1 and the set is bounded. Heine–Borel gives compactness; $\mathrm O(n)=\mathrm U(n)\cap M_n(\mathbb R)$ is a closed subset of a compact set.

Exercise 1.3Components of

\mathrm{GL}(n,\mathbb R)

Show that $\mathrm{GL}^+(n,\mathbb R)$ is connected, so $\mathrm{GL}(n,\mathbb R)$ has exactly two components.

Hint

Gram–Schmidt deforms any $A\in\mathrm{GL}^+$ to an element of $\mathrm{SO}(n)$ through invertible matrices ( $A=QR$ , shrink the upper triangular $R$ to $I$ ). Then join an element of $\mathrm{SO}(n)$ to $I$ by rotating each $2\times2$ block of its normal form.

Solution

Write $A=QR$ with $Q$ orthogonal and $R$ upper triangular with positive diagonal; $\det A>0$ forces $\det Q=1$ . The path $Q((1-t)R+tI)$ stays invertible, since its diagonal stays positive, and ends at $Q\in\mathrm{SO}(n)$ . A rotation is conjugate to a block-diagonal matrix of planar rotations by angles $\theta_j$ (and 1s); scaling all $\theta_j$ to 0 joins it to $I$ inside $\mathrm{SO}(n)$ . The two sets $\det>0$ and $\det<0$ are then the components.

Exercise 1.4Not compact

Show that $\mathrm{SL}(n,\mathbb R)$ for $n\ge2$ , the Heisenberg group and $\mathrm O(1,1)$ are not compact.

Hint

Find an unbounded sequence in each, as in Example 1.4.

Exercise 1.5The dense winding

In Example 1.5, show the set is a subgroup, that it is dense in the torus when $a$ is irrational, and therefore not closed.

Hint

Look at $t=2\pi k$ : the second coordinate $e^{2\pi iak}$ runs over a dense subset of the circle, since $\{ak\bmod1\}$ is dense for irrational $a$ (pigeonhole).

Exercise 1.6The centre of

\mathrm{SU}(2)

Show that the centre of $\mathrm{SU}(2)$ is $\{\pm I\}$ , and that the centre of $\mathrm U(n)$ is the scalar matrices $e^{i\theta}I$ .

Hint

A matrix commuting with every diagonal unitary is diagonal; one that also commutes with permutation matrices is scalar.

Exercise 1.7

\mathrm U(n)

as a space

Show that $U\mapsto(U\,\mathrm{diag}(\det U^{-1},1,\dots,1),\ \det U)$ is a homeomorphism $\mathrm U(n)\to\mathrm{SU}(n)\times\mathrm U(1)$ . Is it a group isomorphism?

Hint

Its inverse is $(V,z)\mapsto V\mathrm{diag}(z,1,\dots,1)$ . It is not a homomorphism for $n\ge2$ ; the group $\mathrm U(n)$ is $(\mathrm{SU}(n)\times\mathrm U(1))/(\mathbb Z/n)$ instead.

Exercise 1.8Heisenberg commutators

Compute the commutator $ghg^{-1}h^{-1}$ of two elements of the Heisenberg group, and show the centre consists of the matrices with $a=b=0$ .

Hint

Using Example 1.7, the commutator has $a=b=0$ and corner entry $ab'-a'b$ .

InteractiveHall Ch. 3 §§1–4

The matrix exponential

The exponential series makes sense for matrices and turns a single matrix $X$ into a whole curve $t\mapsto e^{tX}$ of invertible matrices, a one-parameter subgroup. It is the bridge between linear algebra (the matrix $X$ ) and the group (the curve), and it solves the linear ODE $\dot x=Xx$ .

The idea

Definition (Hall Ch. 3 §1).

e^X=\sum_{k\ge0}X^k/k!

, which converges absolutely because

\lVert X^k\rVert\le\lVert X\rVert^k

. Then

e^0=I

,

(e^X)^{-1}=e^{-X}

,

e^{CXC^{-1}}=Ce^XC^{-1}

, and

e^{X+Y}=e^Xe^Y

when $XY=YX$ . In general it fails (Example 2.5).

Key identities (Hall Ch. 3 §§1, 4).

\frac{d}{dt}e^{tX}=Xe^{tX}=e^{tX}X

, so

x(t)=e^{tX}x_0

solves

\dot x=Xx

. And

\det e^X=e^{\mathrm{tr}X}

(Exercise 2.2), so

e^X

is always invertible, and lies in

\mathrm{SL}

when

\mathrm{tr}X=0

.

Computing it (Hall Ch. 3 §2). If

X=CDC^{-1}

with

D

diagonal,

e^X=Ce^DC^{-1}

. If

X

is nilpotent the series stops. In general

X=S+N

with

S

diagonalizable,

N

nilpotent and

SN=NS

(the Jordan decomposition from the Linear Algebra page), so

e^X=e^Se^N

.

One-parameter subgroups and the logarithm (Hall Ch. 3 §§3–4). Every continuous homomorphism

\mathbb R\to\mathrm{GL}(n,\mathbb C)

is

t\mapsto e^{tX}

for a unique

X

, namely its derivative at 0. Near

I

,

\log A=\sum_{m\ge1}(-1)^{m+1}(A-I)^m/m

inverts the exponential, so

\exp

is a diffeomorphism from a neighbourhood of 0 onto a neighbourhood of

I

.

Explore: one-parameter subgroups of $\mathrm{GL}(2,\mathbb R)$

Curves

t\mapsto e^{tX}p

for several starting points (the phase portrait of

\dot x=Xx

), and the image of the unit circle under

e^{X}

.

Preset X

X₁₁ X₁₂ X₂₁ X₂₂

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Matrix exponential: The power series of the exponential function, fed a square matrix instead of a number; it always converges and always gives an invertible matrix.
One-parameter subgroup: A continuous curve of group elements through the identity with $g(s+t)=g(s)g(t)$ : moving for time $s$ and then $t$ is the same as moving for $s+t$ .
Nilpotent: Describes a matrix some power of which is zero, so that the exponential series stops after finitely many terms.
Matrix logarithm: The inverse of the exponential near the identity, given by the series for $\log(1+x)$ .
Lie product formula: Alternating many tiny steps of two motions approximates moving along their sum: $(e^{X/m}e^{Y/m})^m$ tends to $e^{X+Y}$ .

Examples

Example 2.1 (rotations).

For $X=\begin{pmatrix}0&-\theta\\\theta&0\end{pmatrix}$ , $X^2=-\theta^2I$ , so the series splits into the cosine and sine series: $e^X=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$ .

Example 2.2 (a shear).

$X=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has $X^2=0$ , so $e^{tX}=I+tX=\begin{pmatrix}1&t\\0&1\end{pmatrix}$ : Example 1.3 of the representation theory page, now as a one-parameter subgroup.

Example 2.3 (hyperbolic).

$X=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ has $X^2=I$ , so $e^{tX}=\begin{pmatrix}\cosh t&\sinh t\\\sinh t&\cosh t\end{pmatrix}$ , a Lorentz boost. The orbits are hyperbolas.

Example 2.4 (spirals).

$X=\begin{pmatrix}a&-b\\b&a\end{pmatrix}=aI+bJ$ with $J$ the rotation generator. Since $aI$ and $bJ$ commute, $e^{tX}=e^{at}\begin{pmatrix}\cos bt&-\sin bt\\\sin bt&\cos bt\end{pmatrix}$ : a rotation times a dilation, and $\det e^{tX}=e^{2at}=e^{t\,\mathrm{tr}X}$ .

Example 2.5 (non-commuting).

With $X=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ , $Y=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ : $e^Xe^Y=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ , but $X+Y$ is the matrix of Example 2.3, so $e^{X+Y}=\begin{pmatrix}\cosh1&\sinh1\\\sinh1&\cosh1\end{pmatrix}$ . Module 5 measures the difference.

Example 2.6 (the exponential need not be onto).

$A=\begin{pmatrix}-1&1\\0&-1\end{pmatrix}\in\mathrm{SL}(2,\mathbb R)$ is not $e^X$ for any $X\in\mathfrak{sl}(2,\mathbb R)$ (Exercise 2.6). The group is connected, but its exponential map misses some elements; for compact connected groups it is onto.

Check your understanding

When is $e^{X+Y}=e^Xe^Y$ guaranteed?

When $XY=YX$ .
Always, as for numbers.Example 2.5.
When $X$ and $Y$ are both diagonalizable.They also need to commute; take two non-commuting symmetric matrices.

If they commute, the binomial theorem applies to

(X+Y)^k

and the series multiply as for numbers.

For a real matrix $X$ , $\det e^X$ is:

$e^{\det X}$ Try $X=\mathrm{diag}(1,1)$ : $\det e^X=e^2$ , but $e^{\det X}=e$ .
$e^{\mathrm{tr}X}$ , so always positive.
$\pm1$ Only for trace 0 is it 1.

So the exponential of a real matrix always lands in the identity component

\mathrm{GL}^+(n,\mathbb R)

: one-parameter subgroups start at

I

and can't change the sign of the determinant.

Exercises

Exercise 2.1Convergence and commuting sums

Show that the exponential series converges absolutely for every $X$ , and that $e^{X+Y}=e^Xe^Y$ when $XY=YX$ .

Hint

Compare with $\sum\lVert X\rVert^k/k!$ . For the product, multiply the two absolutely convergent series and regroup by total degree using the binomial theorem.

Solution

$\sum\lVert X^k/k!\rVert\le\sum\lVert X\rVert^k/k!=e^{\lVert X\rVert}$ . Absolutely convergent series may be multiplied and rearranged: $e^Xe^Y=\sum_{j,k}\frac{X^jY^k}{j!k!}=\sum_m\frac1{m!}\sum_{j+k=m}\binom mjX^jY^k=\sum_m\frac{(X+Y)^m}{m!}$ , the last step using $XY=YX$ .

Exercise 2.2

\det e^X=e^{\mathrm{tr}X}

Prove $\det e^X=e^{\mathrm{tr}X}$ .

Hint

Over $\mathbb C$ , $X=CTC^{-1}$ with $T$ upper triangular. Then $e^T$ is upper triangular with diagonal entries $e^{t_{ii}}$ .

Solution

Powers of an upper triangular $T$ are upper triangular with diagonal $t_{ii}^k$ , so $e^T$ is upper triangular with diagonal $e^{t_{ii}}$ and $\det e^T=\prod e^{t_{ii}}=e^{\mathrm{tr}T}$ . Conjugation changes neither side: $\det e^X=\det(Ce^TC^{-1})=\det e^T$ and $\mathrm{tr}X=\mathrm{tr}T$ .

Exercise 2.3Rotation times dilation

Verify Example 2.4, and find all $X$ with $e^{tX}\in\mathrm{SO}(2)$ for all $t$ .

Hint

Differentiate $(e^{tX})^\top e^{tX}=I$ at $t=0$ : $X^\top+X=0$ . Answer: the multiples of $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ .

Exercise 2.4One-parameter subgroups

(1) Show $t\mapsto e^{tX}$ is a smooth homomorphism with derivative $X$ at 0. (2) Show that every continuous homomorphism $A:\mathbb R\to\mathrm{GL}(n,\mathbb C)$ is of this form.

Hint

(2) For small $t$ , $A(t)$ is near $I$ , so $A(t)=e^{Z(t)}$ with $Z(t)=\log A(t)$ small. Fix a small $t_0$ and show $Z(t_0/2^k)=Z(t_0)/2^k$ using uniqueness of small logarithms; then $A(qt_0)=e^{qZ(t_0)}$ for dyadic $q$ , and continuity finishes.

Solution

(1) $e^{(s+t)X}=e^{sX}e^{tX}$ by Exercise 2.1, since $sX$ and $tX$ commute; differentiate the series term by term. (2) Choose $t_0>0$ with $A(t)$ in the neighbourhood where $\log$ is a two-sided inverse of $\exp$ , for $|t|\le t_0$ , and put $X=Z(t_0)/t_0$ . $A(t_0/2)^2=A(t_0)$ , and $e^{Z(t_0/2)}$ squared is $e^{2Z(t_0/2)}$ ; both $2Z(t_0/2)$ and $Z(t_0)$ are small logarithms of $A(t_0)$ , so they are equal. By induction $A(t_0/2^k)=e^{t_0X/2^k}$ , hence $A(t)=e^{tX}$ for all dyadic multiples $t$ of $t_0$ , which are dense; both sides are continuous.

Exercise 2.5The Lie product formula

Show $\lim_{m\to\infty}(e^{X/m}e^{Y/m})^m=e^{X+Y}$ .

Hint

For large $m$ , $e^{X/m}e^{Y/m}=I+\frac{X+Y}m+O(m^{-2})$ , so its logarithm is $\frac{X+Y}m+O(m^{-2})$ . Then $(e^{X/m}e^{Y/m})^m=\exp\big(X+Y+O(1/m)\big)$ .

Exercise 2.6A matrix with no real logarithm of trace zero

Show that $A=\begin{pmatrix}-1&1\\0&-1\end{pmatrix}$ is not $e^X$ for any $X\in\mathfrak{sl}(2,\mathbb R)$ .

Hint

The eigenvalues of $X$ are $\pm\lambda$ , with $\lambda$ real or purely imaginary. Consider the three cases $\lambda$ real nonzero, imaginary nonzero, and zero.

Solution

If $\lambda\ne0$ , $X$ has distinct eigenvalues and is diagonalizable over $\mathbb C$ , so $e^X$ is diagonalizable; $A$ is not. If $\lambda=0$ , $X$ is nilpotent, $e^X=I+X$ , and its eigenvalues are 1, not $-1$ . ( $-A$ is $e^X$ with $X=\begin{pmatrix}0&-1\\0&0\end{pmatrix}$ ; and $A$ itself is a product of two exponentials, as every element of a connected group is.)

Exercise 2.7Exponentials in the Heisenberg algebra

For $X=\begin{pmatrix}0&a&c\\0&0&b\\0&0&0\end{pmatrix}$ , compute $e^X$ and show $\exp$ is a bijection from these matrices onto the Heisenberg group.

Hint

$X^3=0$ , so $e^X=I+X+\frac12X^2$ , whose corner entry is $c+\frac12ab$ . Solve back for $c$ .

Animated lessonLee, Ch. 8: Vector Fields · Ch. 9: Integral Curves and Flows

Flows and the Lie bracket

The Lie bracket $[X,Y]$ is easy to define algebraically and hard to picture. This short animation shows what it measures: follow $X$ , then $Y$ , then go back along $X$ and then $Y$ , and see how far you are from where you started.

The idea

Flow. An integral curve of

X

satisfies

\gamma'(t)=X_{\gamma(t)}

. The flow

\theta_t

moves every point along its integral curve for time

t

.

Lie bracket.

[X,Y]f = X(Yf)-Y(Xf)

. Second derivatives cancel, so this is again a vector field. In coordinates:

[X,Y] = \sum_j \big(X Y^j - Y X^j\big)\pd{}{x^j}.

What it measures. With

\theta

the flow of

X

and

\psi

the flow of

Y

,

\psi_{-t}\circ\theta_{-t}\circ\psi_{t}\circ\theta_{t}(p) = p + t^2\,[X,Y]_p + O(t^3).

So $[X,Y]=0$ exactly when the flows commute.

Watch

Vector fields

Time on each leg, t 0.80

X

Y

gap

t^2[X,Y]_p

· drag p

Two vector fields: $X$ in blue and $Y$ in gold. Pick a start point $p$ .
Flow along $X$ for time $t$ .
Then along $Y$ for time $t$ .
Then backwards along $X$ for time $t$ .
Then backwards along $Y$ . If the flows commuted, we'd be back at $p$ .
The loop doesn't close. The gap (red) is close to $t^2[X,Y]_p$ (green), and the match improves as $t$ shrinks.

Worked example

On $\R^2$ , let $X=-y\,\partial_x+x\,\partial_y$ (rotation) and $Y=x\,\partial_x+y\,\partial_y$ (scaling). Compute $[X,Y]$ , interpret the answer using the flows, then compute $[X,\partial_x]$ .

Compute $XY^j$ and $YX^j$ for each component.

$Y^1=x,\ Y^2=y$ , so $XY^1=X(x)=-y$ and $XY^2=X(y)=x$ .

$X^1=-y,\ X^2=x$ , so $YX^1=Y(-y)=-y$ and $YX^2=Y(x)=x$ .
Assemble $[X,Y]$ .

$[X,Y] = (-y-(-y))\,\partial_x + (x-x)\,\partial_y = 0.$
Explain the answer using the flows.

The flow of $X$ is rotation by angle $t$ , and the flow of $Y$ is scaling by $e^{s}$ . Rotating and scaling commute, $R_t(e^{s}p)=e^{s}R_tp$ , so the four-leg loop always closes. Choose "rotation & scaling" above and check that the gap is zero for every $t$ .
Now replace $Y$ with $\partial_x$ . Compute $[X,\partial_x]$ .

The components of $\partial_x$ are constant, so the $XY^j$ terms vanish. And $\partial_x X^1=\partial_x(-y)=0$ , $\partial_x X^2=\partial_x(x)=1$ . So

$[X,\partial_x] = -\partial_y.$

In the animation ("rotation & translation"), the gap points straight down with length about $t^2$ , at every starting point.

Examples

Example 5.1 (the rotation field).

$X=-y\partial_x+x\partial_y$ has flow $\theta_t=$ rotation by angle $t$ . With $Y=\partial_x$ , $[Y,X]=\partial_y$ : moving along $Y$ , then $X$ , then back, produces a drift in the $y$ direction.

Example 5.2 (coordinate fields commute).

$[\partial_i,\partial_j]=0$ because mixed partials commute. Conversely, two vector fields with zero bracket have commuting flows, and near a point where they are independent they are coordinate fields.

Example 5.3 (matrix groups).

On a matrix Lie group, the left-invariant fields $X^L(g)=gX$ have flows $ge^{tX}$ , and $[X^L,Y^L]=(XY-YX)^L$ : the vector-field bracket is the matrix commutator (Lie Groups, module 4).

Example 5.4 (the Lie derivative).

$[X,Y]=\frac d{dt}\big|_0(\theta_{-t})_*Y$ : the bracket measures how $Y$ changes along the flow of $X$ . It vanishes exactly when $Y$ is invariant under the flow.

Check your understanding

$X=\partial_x$ , $Y=x\,\partial_y$ . What is $[X,Y]$ ?

$\partial_y$
$-\partial_y$ Check the order: $XY^j-YX^j$ .
$0$ $Y$ depends on $x$ , which is the direction $X$ moves in.
$x\,\partial_x$

XY^2=\partial_x(x)=1

and every other term is 0. This is the first pair in the animation. Move

p

around and the gap stays the same, because the bracket is constant.

Which identity is false in general?

$[X,Y]=-[Y,X]$ That one is true (antisymmetry).
$[fX,Y]=f[X,Y]$
$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$ That's the Jacobi identity, which always holds.

The correct rule is

[fX,Y]=f[X,Y]-(Yf)X

. The bracket is not

C^\infty

-linear, so it is not a tensor. It involves derivatives of the coefficients.

On $\R$ , what is the integral curve of $X=x^2\,\partial_x$ starting at $x=1$ ?

$x(t)=e^t$ That solves $x'=x$ , not $x'=x^2$ .
$x(t)=1/(1-t)$ , defined only for $t<1$
$x(t)=1+t^2$

Solving

x'=x^2

gives

x(t)=1/(1-t)

, which runs off to infinity in finite time. So

X

is not complete: its flow is not defined for all

t

.

Which vector fields are guaranteed to be complete?

Compactly supported ones (so every vector field on a compact manifold)
Bounded ones on any manifold $\partial_x$ on the open interval $(0,1)$ is bounded but runs off the end.
Polynomial ones on $\R^n$ The previous question is a counterexample.

Outside the support the flow is stationary, and on the compact support there is a uniform time for which all integral curves exist. That uniform time lets you extend every curve forever.

Practice

Exercise 5.1A hyperbolic flow: solving for a flow

Find the flow of $X=x\,\partial_x-y\,\partial_y$ on $\R^2$ and sketch its integral curves.

Solution

$x'=x,\ y'=-y$ gives $\theta_t(x,y)=(e^tx,\ e^{-t}y)$ . The integral curves are the hyperbolas $xy=\text{const}$ , the four half-axes, and the fixed point at the origin.

Exercise 5.2The bracket with functions: a product rule

Show $[fX,gY]=fg[X,Y]+f(Xg)Y-g(Yf)X$ .

Solution

Apply both sides to $h$ :

\begin{aligned} fX(gYh)-gY(fXh) &= f(Xg)(Yh)+fg\,XYh\\ &\quad -g(Yf)(Xh)-gf\,YXh. \end{aligned}

The two second-order terms combine to $fg[X,Y]h$ .

Exercise 5.3A shear flow: flows that don't commute

Find the flows of $\partial_x$ and $x\,\partial_y$ . Compute the four-leg loop from $(0,0)$ exactly and compare it with $t^2[X,Y]$ .

Solution

$\theta_t(x,y)=(x+t,y)$ and $\psi_t(x,y)=(x,y+tx)$ . From $(0,0)$ : $(t,0)\to(t,t^2)\to(0,t^2)\to(0,t^2)$ . The gap is exactly $t^2\partial_y=t^2[X,Y]$ , with no $O(t^3)$ error, because these fields are so simple.

InteractiveHall Ch. 3 §§5–9

The Lie algebra of a Lie group

The Lie algebra is the tangent space at the identity, with a bracket. In the matrix picture it is the set of $X$ whose one-parameter subgroup stays in the group, and the bracket is the commutator. In the manifold picture it is the left-invariant vector fields, and the bracket is the bracket of vector fields from the previous module. The two pictures agree, which is why Lie theory sits where linear algebra and differential geometry meet.

The idea

Lie algebra of a matrix group (Hall Ch. 3 §5).

\mathfrak g=\{X:e^{tX}\in G\text{ for all }t\in\mathbb R\}

. It is a real vector space closed under

[X,Y]=XY-YX

(Exercise 4.2), and it is the tangent space of

G

at

I

. Examples:

\mathfrak{gl}(n)

is all matrices;

\mathfrak{sl}(n)

is

\mathrm{tr}X=0

;

\mathfrak{so}(n)

is

X^\top=-X

;

\mathfrak u(n)

is

X^*=-X

;

\mathfrak{su}(n)

adds

\mathrm{tr}X=0

; the Heisenberg algebra is the strictly upper triangular

3\times3

matrices.

Homomorphisms differentiate (Hall Ch. 3 §6). A continuous homomorphism

\Phi:G\to H

has a unique linear

\phi:\mathfrak g\to\mathfrak h

with

\Phi(e^X)=e^{\phi(X)}

, and

\phi([X,Y])=[\phi(X),\phi(Y)]

. For

G

acting on itself by conjugation this gives the adjoint representation:

\mathrm{Ad}_g(X)=gXg^{-1}

,

\mathrm{ad}_X(Y)=[X,Y]

, and

\mathrm{Ad}_{e^X}=e^{\mathrm{ad}_X}

(Exercise 4.3).

The manifold picture. For

X\in\mathfrak g

, the vector field

X^L(g)=gX

is left-invariant: left translation by any

h

carries it to itself. Its integral curve through

g

is

t\mapsto ge^{tX}

, so the exponential map is the time-1 flow of

X^L

from the identity. The bracket of the vector fields matches the commutator:

[X^L,Y^L]=[X,Y]^L

, with the vector-field bracket of module 3 (Exercise 4.6). Right-invariant fields

X^R(g)=Xg

have flows

e^{tX}g

and satisfy

[X^R,Y^R]=-[X,Y]^R

.

Generation (Hall Ch. 3 §7). A connected matrix Lie group is generated by

\exp(\mathfrak g)

: every element is a finite product

e^{X_1}\cdots e^{X_k}

(Exercise 4.7), even when

\exp

is not onto (Example 2.6).

Explore: invariant vector fields on the group of $x\mapsto ax+b$

The group of maps

x\mapsto ax+b

with

a>0

is the half-plane of pairs

(a,b)

, as matrices

\begin{pmatrix}a&b\\0&1\end{pmatrix}

. Arrows show the chosen invariant field for

X=\begin{pmatrix}\alpha&\beta\\0&0\end{pmatrix}

; curves are its integral curves. Drag the dot

g

to follow the curve through it.

Field

α β

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Lie algebra of a group: The matrices whose exponential curves stay inside the group: the infinitesimal motions available at the identity, with the commutator as bracket.
Tangent space at the identity: All velocity vectors of smooth curves in the group passing through the identity.
Adjoint representation: The group acting on its own infinitesimal motions by conjugation: $X\mapsto gXg^{-1}$ .
Left-invariant vector field: A field of arrows on the group that looks the same after multiplying every point on the left by a fixed element; it is determined by its arrow at the identity.
Exponential map: Following the flow of an invariant field for unit time, starting at the identity; for matrices, the power series.
Lie algebra homomorphism: A linear map that turns commutators into commutators; differentiating a homomorphism of groups at the identity produces one.

Examples

Example 4.1 (

\mathfrak{so}(3)

is the cross product).

The basis $L_1,L_2,L_3$ of infinitesimal rotations about the axes, $(L_k)_{ij}=-\varepsilon_{kij}$ , satisfies $[L_1,L_2]=L_3$ and cyclically. Identifying $\sum x_kL_k$ with $x\in\mathbb R^3$ , the bracket becomes the cross product and $L_xv=x\times v$ .

Example 4.2 (

\mathfrak{su}(2)\cong\mathfrak{so}(3)

).

$\mathfrak{su}(2)$ has basis $E_k=-\frac i2\sigma_k$ (Pauli matrices), with $[E_1,E_2]=E_3$ cyclically: the same relations as Example 4.1. The Lie algebras are isomorphic although the groups are not; module 6 explains the difference.

Example 4.3 (the affine group).

$\mathfrak g=\Big\{\begin{pmatrix}\alpha&\beta\\0&0\end{pmatrix}\Big\}$ , with $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ , $B=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $[A,B]=B$ : the only non-abelian 2-dimensional Lie algebra. In coordinates $(a,b)$ , the left-invariant field of $X$ is $(a\alpha,a\beta)$ and the right-invariant one is $(\alpha a,\alpha b+\beta)$ ; the figure draws both.

Example 4.4 (the determinant).

$\det:\mathrm{GL}(n)\to\mathrm{GL}(1)$ differentiates to $\mathrm{tr}:\mathfrak{gl}(n)\to\mathbb R$ or $\mathbb C$ , since $\det e^{tX}=e^{t\,\mathrm{tr}X}$ . Kernels match: $\mathrm{SL}(n)$ has Lie algebra $\mathfrak{sl}(n)$ .

Example 4.5 (dimensions).

$\dim\mathfrak{so}(n)=\frac{n(n-1)}2$ , $\dim\mathfrak u(n)=n^2$ , $\dim\mathfrak{su}(n)=n^2-1$ , $\dim\mathfrak{sl}(n,\mathbb R)=n^2-1$ . These are the dimensions of the groups as manifolds.

Example 4.6 (adjoint action of

\mathrm{SU}(2)

).

$\mathrm{Ad}_U(X)=UXU^{-1}$ preserves $\mathfrak{su}(2)\cong\mathbb R^3$ and the inner product $\langle X,Y\rangle=-2\,\mathrm{tr}(XY)$ , so $\mathrm{Ad}$ is a homomorphism $\mathrm{SU}(2)\to\mathrm{SO}(3)$ : the double cover of module 6.

Check your understanding

What is the Lie algebra of $\mathrm{SO}(n)$ ?

Orthogonal matrices.Those are the group, not the tangent space.
Skew-symmetric matrices, $X^\top=-X$ .
Symmetric matrices.Their exponentials are positive definite, not orthogonal.

Differentiate

(e^{tX})^\top e^{tX}=I

at

t=0

(Exercise 4.1).

The integral curve through $g$ of the left-invariant field $X^L(h)=hX$ is:

$t\mapsto ge^{tX}$
$t\mapsto e^{tX}g$ That is the curve of the right-invariant field $Xh$ .
$t\mapsto g+tX$ That leaves the group.

\frac d{dt}ge^{tX}=ge^{tX}X=X^L(ge^{tX})

.

$\dim\mathrm{SU}(3)=$

9That's $\dim\mathrm U(3)$ .
8
3That's $\dim\mathrm{SU}(2)$ .

\mathfrak{su}(3)

is the traceless skew-Hermitian

3\times3

matrices:

9-1=8

real dimensions, the "eightfold way" of module 8.

Exercises

Exercise 4.1

\mathfrak{so}(n)

and

\mathfrak u(n)

Show that the Lie algebra of $\mathrm O(n)$ (and of $\mathrm{SO}(n)$ ) is the skew-symmetric matrices, and that of $\mathrm U(n)$ the skew-Hermitian ones.

Hint

One direction: differentiate. The other: $(e^{tX})^\top=e^{tX^\top}=e^{-tX}$ .

Solution

If $e^{tX}\in\mathrm O(n)$ for all $t$ , differentiating $e^{tX^\top}e^{tX}=I$ at 0 gives $X^\top+X=0$ . Conversely, if $X^\top=-X$ then $(e^{tX})^\top e^{tX}=e^{-tX}e^{tX}=I$ (commuting exponents). Also $\det e^{tX}=e^{t\,\mathrm{tr}X}=1$ , so these land in $\mathrm{SO}(n)$ . The unitary case is the same with $*$ for $\top$ .

Exercise 4.2

\mathfrak g

is a Lie algebra

Show that $\mathfrak g$ is closed under addition (using the Lie product formula, Exercise 2.5) and under $[X,Y]$ .

Hint

$e^{t(X+Y)}=\lim(e^{tX/m}e^{tY/m})^m$ , and $G$ is closed. For the bracket, $e^{tX}Ye^{-tX}\in\mathfrak g$ (why?) and $[X,Y]$ is its derivative at $t=0$ .

Solution

Sum: each $(e^{tX/m}e^{tY/m})^m$ is in $G$ , the limit $e^{t(X+Y)}$ is invertible, and $G$ is closed. Bracket: for $g\in G$ , $e^{t\,gYg^{-1}}=ge^{tY}g^{-1}\in G$ , so $gYg^{-1}\in\mathfrak g$ . Hence $c(t)=e^{tX}Ye^{-tX}$ is a curve in the vector space $\mathfrak g$ , which is closed in the space of matrices, so its derivative $c'(0)=XY-YX$ is in $\mathfrak g$ .

Exercise 4.3

\mathrm{Ad}_{e^X}=e^{\mathrm{ad}_X}

Prove $e^XYe^{-X}=e^{\mathrm{ad}_X}(Y)=Y+[X,Y]+\frac1{2!}[X,[X,Y]]+\dots$

Hint

Both $A(t)=\mathrm{Ad}_{e^{tX}}$ and $B(t)=e^{t\,\mathrm{ad}_X}$ solve $\dot Z=\mathrm{ad}_X\circ Z$ , $Z(0)=\mathrm{id}$ .

Solution

$\frac d{dt}e^{tX}Ye^{-tX}=e^{tX}(XY-YX)e^{-tX}=\mathrm{ad}_X(e^{tX}Ye^{-tX})$ , using that $X$ commutes with $e^{tX}$ . So $A(t)Y$ solves the linear ODE $\dot Z=\mathrm{ad}_XZ$ with initial value $Y$ , whose unique solution is $e^{t\,\mathrm{ad}_X}Y$ . Set $t=1$ .

Exercise 4.4

\mathfrak{so}(3)

and the cross product

Verify Example 4.1: write out $L_1,L_2,L_3$ , compute their brackets, and show $L_xv=x\times v$ .

Hint

$L_3=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}$ generates rotations about the $z$ -axis. The Jacobi identity for the bracket becomes the Jacobi identity for the cross product.

Exercise 4.5Left-invariant fields and their flows

For a matrix Lie group, show that $X^L(g)=gX$ is invariant under left translations $L_h(g)=hg$ , and that its flow is $\theta_t(g)=ge^{tX}$ . Deduce that left-invariant fields are complete, and that $\exp(X)=\theta_1(I)$ .

Hint

The differential of $L_h$ is $v\mapsto hv$ (it is linear on matrices).

Solution

$dL_h(X^L(g))=hgX=X^L(hg)$ , which is invariance. $\frac d{dt}ge^{tX}=ge^{tX}X=X^L(ge^{tX})$ with value $g$ at $t=0$ , so by uniqueness of integral curves this is the flow; it is defined for all $t$ , so the field is complete. At $g=I$ and $t=1$ it gives $e^X$ .

Exercise 4.6The two brackets agree

On the affine group of Example 4.3, compute the left-invariant fields $A^L,B^L$ in coordinates $(a,b)$ and the vector-field bracket $[A^L,B^L]$ . Check that it equals $[A,B]^L=B^L$ . Do the same for the right-invariant fields and find the sign.

Hint

$A^L=a\partial_a$ , $B^L=a\partial_b$ ; $A^R=a\partial_a+b\partial_b$ , $B^R=\partial_b$ . With $[V,W]f=V(Wf)-W(Vf)$ : $[a\partial_a,a\partial_b]=a\partial_b$ , while $[A^R,B^R]=-\partial_b=-[A,B]^R$ .

Exercise 4.7Exponentials generate

Show that if $G$ is connected, every element is a product of finitely many exponentials.

Hint

The subgroup generated by $\exp(\mathfrak g)$ contains a neighbourhood of $I$ (the logarithm), so it is open; its complement is a union of cosets, also open. Connectedness does the rest.

Exercise 4.8An explicit isomorphism

\mathfrak{su}(2)\cong\mathfrak{so}(3)

Show that $E_k\mapsto L_k$ (Examples 4.1 and 4.2) is a Lie algebra isomorphism, and that it is the derivative of $\mathrm{Ad}:\mathrm{SU}(2)\to\mathrm{SO}(3)$ in suitable bases.

Hint

Both bases satisfy $[X_1,X_2]=X_3$ cyclically. The derivative of $\mathrm{Ad}$ is $\mathrm{ad}$ , and $\mathrm{ad}_{E_1}$ has matrix $L_1$ in the basis $E_1,E_2,E_3$ .

InteractiveHall Ch. 4 §§1–4

The Baker–Campbell–Hausdorff formula

Multiplying $e^X$ by $e^Y$ gives $e^Z$ for some $Z$ , at least near the identity. The remarkable fact is that $Z$ can be written using only $X$ , $Y$ and brackets. So the Lie algebra, with nothing but its bracket, determines the group multiplication near the identity.

The idea

BCH (Hall Ch. 4 §§2–3). For

X,Y

small enough,

e^Xe^Y=e^Z

with

Z=X+Y+\tfrac12[X,Y]+\tfrac1{12}\big([X,[X,Y]]-[Y,[X,Y]]\big)+\cdots,

where every further term is an iterated bracket of

X

and

Y

. Hall proves the integral form

Z=X+\int_0^1g(e^{\mathrm{ad}_X}e^{t\,\mathrm{ad}_Y})(Y)\,dt

with

g(z)=\frac{z\log z}{z-1}

.

The Heisenberg case (Hall Ch. 4 §1). If

[X,Y]

commutes with both

X

and

Y

, the series stops:

e^Xe^Y=e^{X+Y+\frac12[X,Y]}

exactly (Exercise 5.4). This holds in the Heisenberg algebra, and it is the formula behind the canonical commutation relations of quantum mechanics.

The Lie correspondence (Hall Ch. 4 §4, Ch. 5 §8). Consequences of BCH: every Lie subalgebra

\mathfrak h\subseteq\mathfrak g

is the Lie algebra of a unique connected subgroup; and if

G

is simply connected, every Lie algebra homomorphism

\mathfrak g\to\mathfrak h

is the derivative of a unique group homomorphism

G\to H

. Simple connectivity cannot be dropped (Exercise 5.6).

Explore: how good is each truncation?

Error

\lVert\log(e^{sX}e^{sY})-Z_k(s)\rVert

against

s

, on log–log axes, for the truncation after degree 1 , degree 2 and degree 3 . Slopes 2, 3, 4 mean errors of order

s^2,s^3,s^4

.

X and Y

s (marked on the plot)

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Baker–Campbell–Hausdorff formula: The recipe for the single exponent that equals doing one exponential and then another, written entirely in terms of the two exponents and their commutators.
Group commutator: Doing $g$ , then $h$ , then undoing $g$ , then undoing $h$ . It is the identity exactly when $g$ and $h$ commute, and for small motions it is approximately the exponential of the bracket.
Lie correspondence: The dictionary between groups and their infinitesimal versions: subalgebras give subgroups, and maps of algebras give maps of groups when the source group has no non-shrinkable loops.
Locally isomorphic: Describes two groups that look the same near the identity, such as the line and the circle, or $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ , because their Lie algebras are isomorphic.

Examples

Example 5.1.

If $X$ and $Y$ commute, every bracket vanishes and $Z=X+Y$ , recovering Exercise 2.1.

Example 5.2 (Heisenberg).

For $X=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}$ , $Y=\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}$ , $[X,Y]=E_{13}$ is central, and $e^{sX}e^{sY}=e^{s(X+Y)+\frac{s^2}2E_{13}}$ exactly: in the figure, the degree-2 and degree-3 errors are at rounding level.

Example 5.3 (parallel parking with rotations).

Small rotations about the $x$ - and $y$ -axes compose to a rotation about $x+y$ , corrected by $\frac12[L_1,L_2]=\frac12L_3$ : a small turn about the $z$ -axis appears from nowhere. The group commutator $e^{sL_1}e^{sL_2}e^{-sL_1}e^{-sL_2}=e^{s^2L_3+O(s^3)}$ makes it visible (Exercise 5.3).

Example 5.4 (same algebra, different groups).

$\mathbb R$ and $\mathrm{SO}(2)$ both have Lie algebra $\mathbb R$ with zero bracket. BCH holds in both, but it only describes the group near the identity: $\theta\mapsto e^{i\theta}$ wraps the line around the circle.

Example 5.5 (

\mathfrak{sl}(2)

).

For $e$ and $f$ , $[e,f]=h$ , and all the higher brackets are nonzero, so no truncation of the series is exact. In the figure each truncation gains one power of $s$ .

Check your understanding

To what order in $s$ does $e^{sX}e^{sY}$ differ from $e^{s(X+Y)}$ ?

$s$ The first-order terms agree.
$s^2$ , with leading term $\frac{s^2}2[X,Y]$ .
$s^3$ Only when $[X,Y]=0$ .

The red line in the figure has slope 2, except for commuting

X,Y

.

The Lie algebra homomorphism $\mathfrak{so}(3)\to\mathfrak{su}(2)$ inverse to the derivative of the double cover:

comes from a group homomorphism $\mathrm{SO}(3)\to\mathrm{SU}(2)$ .That homomorphism would split the double cover.
comes from no group homomorphism, because $\mathrm{SO}(3)$ is not simply connected.

Exercise 5.6. This is exactly why the spin representations of

\mathrm{SU}(2)

are not representations of

\mathrm{SO}(3)

.

Exercises

Exercise 5.1BCH to second order

Using $\log(I+A)=A-\frac12A^2+O(A^3)$ , show $\log(e^{sX}e^{sY})=s(X+Y)+\frac{s^2}2[X,Y]+O(s^3)$ .

Hint

$e^{sX}e^{sY}=I+A$ with $A=s(X+Y)+s^2(\frac12X^2+XY+\frac12Y^2)+O(s^3)$ .

Solution

$A^2=s^2(X+Y)^2+O(s^3)$ , so $\log(I+A)=s(X+Y)+s^2\big(\frac12X^2+XY+\frac12Y^2-\frac12(X^2+XY+YX+Y^2)\big)+O(s^3)=s(X+Y)+\frac{s^2}2(XY-YX)+O(s^3)$ .

Exercise 5.2The Heisenberg group multiplication

Using Exercise 2.7, verify Example 5.2 directly for arbitrary elements of the Heisenberg algebra: $e^Xe^Y=e^{X+Y+\frac12[X,Y]}$ .

Hint

Write $X=(a,b,c)$ for the entries above the diagonal. Then $[X,Y]=(0,0,ab'-a'b)$ , and both sides have corner entry $c+c'+\frac12(a+a')(b+b')+\frac12(ab'-a'b)=c+c'+\frac12ab+\frac12a'b'+ab'$ .

Solution

By Exercise 2.7, $e^X$ is the group element with entries $a,b$ and corner $c+\frac12ab$ . By Example 1.7, the product $e^Xe^Y$ has entries $a+a',b+b'$ and corner $c+\frac12ab+c'+\frac12a'b'+ab'$ . The exponent $W=X+Y+\frac12[X,Y]$ has entries $a+a',b+b'$ and corner $c+c'+\frac12(ab'-a'b)$ , so $e^W$ has corner $c+c'+\frac12(ab'-a'b)+\frac12(a+a')(b+b')$ , which expands to the same thing.

Exercise 5.3The group commutator

Show $e^{sX}e^{sY}e^{-sX}e^{-sY}=e^{s^2[X,Y]+O(s^3)}$ .

Hint

Apply Exercise 5.1 twice: $e^{sX}e^{sY}=e^{s(X+Y)+\frac{s^2}2[X,Y]+O(s^3)}$ and $e^{-sX}e^{-sY}=e^{-s(X+Y)+\frac{s^2}2[X,Y]+O(s^3)}$ , then once more.

Exercise 5.4When the series stops

Suppose $C=[X,Y]$ commutes with $X$ and $Y$ . Show $e^{tX}e^{tY}=e^{t(X+Y)+\frac{t^2}2C}$ .

Hint

Show $e^{tX}Ye^{-tX}=Y+tC$ (Exercise 4.3). Then both sides of the claim solve $\dot A=A\,(X+Y+tC)$ with $A(0)=I$ .

Solution

$\mathrm{ad}_X^2Y=[X,C]=0$ , so $e^{tX}Ye^{-tX}=Y+tC$ . For $A(t)=e^{tX}e^{tY}$ : $\dot A=e^{tX}Xe^{tY}+e^{tX}e^{tY}Y=A(e^{-tY}Xe^{tY}+Y)=A(X+tC+Y)$ , using $e^{-tY}Xe^{tY}=X-t[Y,X]=X+tC$ . For $B(t)=e^{t(X+Y)+t^2C/2}$ , the exponent commutes with its derivative (because $C$ is central in the algebra generated by $X,Y$ ), so $\dot B=B(X+Y+tC)$ . Same linear ODE and initial value, so $A=B$ .

Exercise 5.5The line and the circle

Show that $\mathbb R$ and $\mathrm{SO}(2)$ have isomorphic Lie algebras but are not isomorphic as groups.

Hint

One is compact and the other isn't; or, $\mathrm{SO}(2)$ has elements of finite order and $\mathbb R$ has none except 0.

Exercise 5.6Why simple connectivity is needed

Let $\phi:\mathfrak{so}(3)\to\mathfrak{su}(2)$ be the inverse of the isomorphism in Exercise 4.8. Show there is no continuous homomorphism $\Phi:\mathrm{SO}(3)\to\mathrm{SU}(2)$ with derivative $\phi$ .

Hint

Rotation by $2\pi$ about the $z$ -axis is $e^{2\pi L_3}=I$ . What is $e^{2\pi\phi(L_3)}$ ?

Solution

$\phi(L_3)=E_3=-\frac i2\sigma_3$ , so $e^{2\pi E_3}=\mathrm{diag}(e^{-i\pi},e^{i\pi})=-I$ . A homomorphism with derivative $\phi$ would satisfy $\Phi(e^{2\pi L_3})=e^{2\pi\phi(L_3)}$ , i.e. $\Phi(I)=-I$ , which is impossible.

Exercise 5.7The third-order term

Extend Exercise 5.1 to order $s^3$ and recover the coefficient $\frac1{12}$ of $[X,[X,Y]]-[Y,[X,Y]]$ . Check it numerically with the $\mathfrak{sl}(2)$ preset of the figure, where the teal line has slope 4.

Hint

You need $\log(I+A)=A-\frac12A^2+\frac13A^3+O(A^4)$ and the cubic terms $\frac16X^3+\frac12X^2Y+\frac12XY^2+\frac16Y^3$ of $e^{sX}e^{sY}$ . Expect patience rather than ideas.

InteractiveHall Ch. 2 §§5–6, Ch. 5 §9

SU(2), SO(3) and covering groups

Unit quaternions rotate three-dimensional space, and $q$ and $-q$ give the same rotation. This two-to-one map $\mathrm{SU}(2)\to\mathrm{SO}(3)$ is the smallest example of a covering group, and it is the reason a rotation by $360^\circ$ is not "the same as doing nothing" for an electron.

The idea

The double cover (Hall Ch. 2 §6, Ch. 5 §9).

\mathrm{SU}(2)

is the group of unit quaternions

q=a+bi+cj+dk

. Conjugation

v\mapsto qv\bar q

preserves the imaginary quaternions

\mathbb R^3=\{bi+cj+dk\}

and their length, so it defines

\Phi(q)\in\mathrm{SO}(3)

. The map

\Phi

is a smooth surjective homomorphism with kernel

\{\pm1\}

, and it is a local diffeomorphism. The rotation by angle

\theta

about the unit vector

u

is

\Phi\big(\cos\tfrac\theta2+u\sin\tfrac\theta2\big)

.

Covering groups (Hall Ch. 5 §9). A covering homomorphism

\tilde G\to G

is a surjective homomorphism that is a local diffeomorphism; the Lie algebras are then isomorphic. Every connected Lie group has a simply connected universal cover

\tilde G

, and

\pi_1(G)

is the kernel.

\mathrm{SU}(2)\cong S^3

is simply connected, so

\pi_1(\mathrm{SO}(3))=\{\pm1\}\cong\mathbb Z/2

.

Why it matters for representations. Representations of

\mathfrak{so}(3)\cong\mathfrak{su}(2)

all integrate to

\mathrm{SU}(2)

, but only those where

-1

acts trivially descend to

\mathrm{SO}(3)

:

V_m

with

m

even (module 7). The others are the half-integer spins, which are honest representations of

\mathrm{SU}(2)

and only "projective" representations of

\mathrm{SO}(3)

.

Explore: turning twice

Top: a frame rotated by

\theta

about the axis

u

. Bottom: the great circle in

\mathrm{SU}(2)

through

1

and

-1

traced by the lift

q(\theta)=\cos\frac\theta2+u\sin\frac\theta2

. At

\theta=2\pi

the frame is back but the lift is at

-1

; it returns to

1

at

4\pi

.

Angle θ Axis tilt from z

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Unit quaternion: A number $a+bi+cj+dk$ of length 1 in the four-dimensional number system with $i^2=j^2=k^2=ijk=-1$ ; these form a group shaped like a 3-sphere.
Covering homomorphism: A map of groups that is onto and looks like an isomorphism near every point, but may be many-to-one overall, like wrapping a line around a circle.
Universal cover: The group with no non-shrinkable loops that sits above a given connected group and maps onto it locally isomorphically; every other such group sits in between.
Fundamental group: The group of loops based at a point, up to continuous deformation, with "go around one loop then the other" as the operation.
Spin: For an irreducible piece of the 2 by 2 special unitary group, half of its dimension minus one; half-integer values are the ones that do not come from rotations of space.

Examples

Example 6.1 (the line covers the circle).

$t\mapsto e^{it}$ , $\mathbb R\to\mathrm U(1)$ , is a covering homomorphism with kernel $2\pi\mathbb Z\cong\pi_1(\mathrm U(1))$ .

Example 6.2 (a quarter turn).

$q=\cos\frac\pi4+k\sin\frac\pi4$ gives $qi\bar q=j$ and $qj\bar q=-i$ , $qk\bar q=k$ : rotation by $90^\circ$ about the $z$ -axis. So does $-q$ .

Example 6.3 (a loop that doesn't shrink).

$\theta\mapsto R_z(\theta)$ , $0\le\theta\le2\pi$ , is a loop in $\mathrm{SO}(3)$ . Its lift starts at $1$ and ends at $-1$ , so it is not a loop in $\mathrm{SU}(2)$ , and the original loop cannot be shrunk (Exercise 6.4). Going around twice lifts to a closed loop in the simply connected $\mathrm{SU}(2)$ , which does shrink: the "belt trick".

Example 6.4 (

\mathrm{SO}(3)

as a space).

Each rotation is "angle $\theta\in[0,\pi]$ about axis $u$ ", i.e. a point $\theta u$ of the closed ball of radius $\pi$ , with $\pi u$ and $-\pi u$ the same rotation. So $\mathrm{SO}(3)$ is the ball with antipodal boundary points glued, which is $\mathbb{RP}^3$ .

Example 6.5 (

\mathrm{SO}(4)

).

Pairs of unit quaternions act on $\mathbb H=\mathbb R^4$ by $v\mapsto pv\bar q$ , giving $\mathrm{SU}(2)\times\mathrm{SU}(2)\to\mathrm{SO}(4)$ with kernel $\{(1,1),(-1,-1)\}$ . So $\mathfrak{so}(4)\cong\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ .

Example 6.6 (a cover that isn't a matrix group).

$\pi_1(\mathrm{SL}(2,\mathbb R))=\mathbb Z$ (it deformation retracts onto $\mathrm{SO}(2)$ ). Its universal cover is a Lie group with no faithful finite-dimensional representation, so it is not isomorphic to any matrix Lie group.

Check your understanding

What is the kernel of $\Phi:\mathrm{SU}(2)\to\mathrm{SO}(3)$ ?

$\{\pm I\}$
$\{I\}$ Then $\Phi$ would be an isomorphism, but $\mathrm{SU}(2)$ is simply connected and $\mathrm{SO}(3)$ is not.
The diagonal matrices. $\mathrm{diag}(e^{i\phi},e^{-i\phi})$ is a rotation about the $z$ -axis by $2\phi$ , nontrivial unless $e^{i\phi}=\pm1$ .

qv\bar q=v

for all imaginary

v

forces

q

to be real, so

q=\pm1

.

In the figure, at $\theta=2\pi$ the lift $q$ is:

$1$ The frame is back where it started, but look at the right panel.
$-1$

\cos\pi+u\sin\pi=-1

. A full turn is a non-trivial loop in

\mathrm{SO}(3)

, which is the Exercise 5.6 obstruction seen geometrically.

Exercises

Exercise 6.1Quaternions act by rotations

Show that for a unit quaternion $q$ , $v\mapsto qv\bar q$ maps imaginary quaternions to imaginary quaternions, preserves length, and has determinant 1. Show that $\Phi$ is a homomorphism.

Hint

$v$ is imaginary iff $\bar v=-v$ . $|qv\bar q|=|q||v||\bar q|$ . For the determinant, use connectedness of $\mathrm{SU}(2)$ and continuity of $\det$ .

Solution

$\overline{qv\bar q}=q\bar v\bar q=-qv\bar q$ , so the image is imaginary. Quaternion length is multiplicative, so length is preserved, and $\Phi(q)\in\mathrm O(3)$ . $\det\Phi(q)=\pm1$ depends continuously on $q$ on the connected $S^3$ and is 1 at $q=1$ , so it is always 1. $\Phi(pq)v=pqv\bar q\bar p=\Phi(p)\Phi(q)v$ .

Exercise 6.2The angle is doubled

For $q=\cos\frac\theta2+k\sin\frac\theta2$ , show $\Phi(q)$ is rotation by $\theta$ about the $z$ -axis.

Hint

$k$ commutes with $k$ ; compute $qi\bar q$ using $ki=j$ and $ik=-j$ .

Solution

With $c=\cos\frac\theta2$ , $s=\sin\frac\theta2$ : $qi\bar q=(c+sk)i(c-sk)=(ci+sj)(c-sk)=c^2i-cs\,ik+cs\,j-s^2jk=(c^2-s^2)i+2cs\,j=\cos\theta\,i+\sin\theta\,j$ . Similarly $qj\bar q=-\sin\theta\,i+\cos\theta\,j$ , and $qk\bar q=k$ .

Exercise 6.3The kernel and surjectivity

Show $\ker\Phi=\{\pm1\}$ and that $\Phi$ is onto.

Hint

Kernel: $q$ commutes with $i$ , $j$ , $k$ , so it is real. Onto: every rotation is a rotation by some angle about some axis (Euler), and Exercise 6.2 generalizes to any unit axis $u$ .

Exercise 6.4

\pi_1(\mathrm{SO}(3))\ne0

Using the path-lifting property of coverings, show that the loop of Example 6.3 is not null-homotopic in $\mathrm{SO}(3)$ .

Hint

A homotopy of loops lifts to a homotopy of paths starting at $1$ , and the endpoint of the lift varies continuously in the discrete set $\{\pm1\}$ .

Solution

The lift of $\theta\mapsto R_z(\theta)$ starting at $1$ is $\theta\mapsto\cos\frac\theta2+k\sin\frac\theta2$ , which ends at $-1$ . If the loop were homotopic to the constant loop, lifting the homotopy would give a continuous family of lifted paths from $1$ , whose endpoints lie in $\Phi^{-1}(I)=\{\pm1\}$ and so are constant. The constant loop lifts to a path ending at $1$ , a contradiction.

Exercise 6.5

\mathrm{SU}(2)

is simply connected

Show that $S^3$ is simply connected.

Hint

A loop can be perturbed to a smooth loop, which misses some point because its image has measure zero. $S^3$ minus a point is homeomorphic to $\mathbb R^3$ by stereographic projection.

Exercise 6.6

\mathrm{SO}(3)\cong\mathbb{RP}^3

Make Example 6.4 precise: show $\Phi$ induces a homeomorphism $S^3/\{\pm1\}\to\mathrm{SO}(3)$ , and that $S^3/\{\pm1\}=\mathbb{RP}^3$ .

Hint

A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

Exercise 6.7

\mathrm{SO}(4)

Verify Example 6.5: the map is a homomorphism into $\mathrm{SO}(4)$ , with the stated kernel.

Hint

If $pv\bar q=v$ for all $v$ , take $v=1$ to get $p=q$ , then $p$ is central, so $p=q=\pm1$ . Surjectivity follows from a dimension count and connectedness.

InteractiveEtingof et al. §§1.9, 1.13–1.14; Hall Ch. 5 §§4, 6, 8–9

sl(2) and SU(2): the first Lie example

A Lie group is too big to work with element by element, so you differentiate: a representation of the group gives a representation of its Lie algebra, a few matrices satisfying commutation relations. For $\mathfrak{sl}(2)$ the whole classification follows from one observation, that $e$ and $f$ move weights up and down by 2.

The idea

From groups to Lie algebras (Hall Ch. 5 §§1, 8). If

\Pi:G\to\mathrm{GL}(V)

is a smooth representation of a matrix Lie group, then

\pi(X)=\frac{d}{dt}\Pi(e^{tX})\big|_{t=0}

is a representation of its Lie algebra: linear, with

\pi([X,Y])=[\pi(X),\pi(Y)]

. If

G

is connected,

\Pi

is determined by

\pi

; if

G

is also simply connected, every

\pi

comes from some

\Pi

.

\mathrm{SU}(2)

is simply connected and

\mathfrak{su}(2)\otimes\mathbb C=\mathfrak{sl}(2,\mathbb C)

, so its representations are the representations of

\mathfrak{sl}(2)

.

$\mathfrak{sl}(2)$ (Etingof §1.9). Trace-zero

2\times2

matrices, with basis

e=\begin{pmatrix}0&1\\0&0\end{pmatrix}

,

f=\begin{pmatrix}0&0\\1&0\end{pmatrix}

,

h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}

and relations

[h,e]=2e

,

[h,f]=-2f

,

[e,f]=h

. A representation is three matrices satisfying these relations.

Weights move by 2. If

hv=\lambda v

then

h(ev)=(\lambda+2)ev

and

h(fv)=(\lambda-2)fv

(Exercise 7.1). In a finite-dimensional representation

h

has finitely many eigenvalues, so some eigenvector

v

has

ev=0

: a highest weight vector.

Classification (Etingof Thm. 1.1, §1.14; Hall Ch. 5 §4). For each

m=0,1,2,\dots

there is exactly one irreducible representation

V_m

of dimension

m+1

. It has basis

v_k=f^kv_0

,

k=0,\dots,m

, with

hv_k=(m-2k)v_k,\qquad fv_k=v_{k+1},\qquad ev_k=k(m-k+1)v_{k-1}.

Realized on homogeneous polynomials of degree

m

in

x,y

:

e=x\partial_y

,

f=y\partial_x

,

h=x\partial_x-y\partial_y

. Every finite-dimensional representation is a direct sum of these (Weyl; for

\mathrm{SU}(2)

, Hall's averaging over the compact group).

Clebsch–Gordan (Hall Ch. 5 §6).

V_m\otimes V_n\cong V_{m+n}\oplus V_{m+n-2}\oplus\dots\oplus V_{|m-n|}

. Weights of a tensor product add, so you can peel off highest weights one at a time (Exercise 7.4).

Group versus Lie algebra (Hall Ch. 5 §9).

\mathrm{SU}(2)\to\mathrm{SO}(3)

is a double cover with kernel

\pm I

, and

-I=\exp(\pi i h)

acts on

V_m

by

(-1)^m

. So only

V_m

with

m

even are representations of

\mathrm{SO}(3)

: in physics language, integer spin

m/2

. The half-integer spins exist only for

\mathrm{SU}(2)

.

Explore: weight diagrams and Clebsch–Gordan

m (first factor) n (second factor; 0 = no tensor)

Each dot is a weight vector, placed at its eigenvalue of

h

. With

n=0

the arrows show

e

(up by 2) and

f

(down by 2) with their coefficients. With

n>0

, the weights of

V_m\otimes V_n

are stacked and coloured by the irreducible piece they belong to.

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Lie algebra: The infinitesimal version of a continuous group: the space of directions you can move in from the identity, with a bracket recording how badly two such motions fail to commute.
Weight: An eigenvalue of the diagonal element $h$ on a vector; the label that the raising and lowering operators shift by 2.
Highest weight vector: A vector that the raising operator kills. Every irreducible piece is generated from one by applying the lowering operator repeatedly.
Raising and lowering operators: The two off-diagonal elements, which move a vector up or down the ladder of eigenvalues by two steps.
Clebsch–Gordan decomposition: The rule for splitting the product of two irreducible pieces: dimensions $m+1$ and $n+1$ give one piece of each size from $m+n+1$ down to $|m-n|+1$ , in steps of two.
Casimir element: A particular quadratic combination of the basis elements that commutes with everything, so it acts on each irreducible piece by a single number that tells the pieces apart.
Double cover: A two-to-one map of groups that is locally an isomorphism, like the map from unit quaternions to rotations of space, under which $q$ and $-q$ give the same rotation.

Examples

Example 7.1.

$V_0$ is the trivial representation and $V_1=\mathbb C^2$ is the defining one, with weights $\pm1$ on $e_1,e_2$ .

Example 7.2 (the adjoint representation).

$\mathfrak{sl}(2)$ acts on itself by $\mathrm{ad}(X)Y=[X,Y]$ . The weights of $h$ on $e,h,f$ are $2,0,-2$ , and $e$ is a highest weight vector, so this is $V_2$ . Under $\mathrm{SU}(2)\to\mathrm{SO}(3)$ it is the rotation representation on $\mathbb R^3\otimes\mathbb C$ .

Example 7.3 (polynomials).

On degree-2 polynomials with basis $x^2,xy,y^2$ : $h$ has weights $2,0,-2$ ; $f=y\partial_x$ sends $x^2\mapsto2xy\mapsto2y^2$ ; $e=x\partial_y$ sends $y^2\mapsto2xy\mapsto2x^2$ . Up to rescaling the basis, these are the formulas for $V_2$ .

Example 7.4 (two spins).

$V_1\otimes V_1$ has weights $2,0,0,-2$ , so it is $V_2\oplus V_0$ : the symmetric tensors (triplet) and the antisymmetric one $e_1\otimes e_2-e_2\otimes e_1$ (singlet).

Example 7.5 (the Casimir).

$C=ef+fe+\frac12h^2$ commutes with $e,f,h$ (Exercise 7.5), so it is a scalar on each $V_m$ . On $v_0$ : $efv_0=[e,f]v_0+fev_0=mv_0$ and $fev_0=0$ , so $C=m+\frac{m^2}2=\frac{m(m+2)}2$ . In physics this is $2j(j+1)$ for spin $j=m/2$ .

Example 7.6 (spherical harmonics).

$\mathrm{SO}(3)$ acts on homogeneous harmonic polynomials of degree $\ell$ in $x,y,z$ , a space of dimension $2\ell+1$ . It is irreducible, isomorphic to $V_{2\ell}$ . Restricted to the sphere these are the spherical harmonics, and the decomposition of functions on the sphere into them is the $\mathrm{SO}(3)$ analogue of Fourier series.

Check your understanding

What are the weights of $V_3$ ?

$0,1,2,3$ Weights are symmetric about 0.
$3,1,-1,-3$
$3,2,1,0,-1,-2,-3$ Weights move in steps of 2, and $\dim V_3=4$ .

m-2k

for

k=0,\dots,m

.

$V_2\otimes V_1\cong$

$V_3\oplus V_1$
$V_3\oplus V_2\oplus V_1$ Dimensions: $3\cdot2=6$ , but $4+3+2=9$ .
$V_2\oplus V_1\oplus V_0$ The highest weight is $2+1=3$ .

Clebsch–Gordan with

m=2

,

n=1

:

V_3\oplus V_1

, dimensions

4+2=6

.

Why is $V_1$ not a representation of $\mathrm{SO}(3)$ ?

$-I\in\mathrm{SU}(2)$ maps to the identity rotation but acts on $V_1$ by $-1$ .
It is not irreducible.It is irreducible.
Its dimension is even.Dimension isn't the issue by itself; the kernel of $\mathrm{SU}(2)\to\mathrm{SO}(3)$ is.

A representation of

\mathrm{SU}(2)

factors through

\mathrm{SO}(3)

exactly when

-I

acts trivially, i.e.

m

even.

Exercises

Exercise 7.1Raising and lowering

Show that if $hv=\lambda v$ , then $h(ev)=(\lambda+2)ev$ and $h(fv)=(\lambda-2)fv$ .

Hint

$hev=[h,e]v+ehv$ .

Solution

$h(ev)=[h,e]v+e(hv)=2ev+\lambda ev$ . Similarly $h(fv)=-2fv+\lambda fv$ .

Exercise 7.2Classify the irreducibles of

\mathfrak{sl}(2)

Let $V$ be a finite-dimensional irreducible representation. Take a highest weight vector $v_0$ of weight $\lambda$ and put $v_k=f^kv_0$ . Show $ev_k=k(\lambda-k+1)v_{k-1}$ , that the nonzero $v_k$ span $V$ , and that $\lambda=m$ is a non-negative integer with $\dim V=m+1$ .

Hint

Induct on $k$ with $ev_k=efv_{k-1}=[e,f]v_{k-1}+fev_{k-1}$ . Let $m$ be the last index with $v_m\ne0$ and apply the formula to $v_{m+1}=0$ .

Solution

$ev_k=hv_{k-1}+fev_{k-1}=(\lambda-2k+2)v_{k-1}+(k-1)(\lambda-k+2)v_{k-1}=k(\lambda-k+1)v_{k-1}$ . The nonzero $v_k$ have distinct weights, so they are independent, and their span is invariant under $e,f,h$ ; by irreducibility it is $V$ . If $v_m\ne0$ and $v_{m+1}=0$ , then $0=ev_{m+1}=(m+1)(\lambda-m)v_m$ , so $\lambda=m$ . Hence $\dim V=m+1$ , and all the formulas are forced, which also proves uniqueness. Existence: the polynomial model.

Exercise 7.3The polynomial model

Check that $e=x\partial_y$ , $f=y\partial_x$ , $h=x\partial_x-y\partial_y$ satisfy the $\mathfrak{sl}(2)$ relations on polynomials in $x,y$ , and that degree- $m$ polynomials form a copy of $V_m$ .

Hint

Compute on monomials: $h(x^ay^b)=(a-b)x^ay^b$ , $e(x^ay^b)=b\,x^{a+1}y^{b-1}$ . The monomial $x^m$ is a highest weight vector of weight $m$ .

Exercise 7.4Prove Clebsch–Gordan

Assuming complete reducibility, prove $V_m\otimes V_n\cong\bigoplus_{k=0}^{\min(m,n)}V_{m+n-2k}$ .

Hint

The weights of a tensor product are the sums of weights. A representation is determined by its weight multiplicities, since $V_j$ contributes one vector to each weight $j,j-2,\dots,-j$ .

Solution

$h$ acts on $v_i\otimes w_j$ by $(m-2i)+(n-2j)$ . The multiplicity of weight $m+n-2s$ is the number of $(i,j)$ with $i+j=s$ , $0\le i\le m$ , $0\le j\le n$ : it increases by 1 from $s=0$ up to $s=\min(m,n)$ , then stays constant until the symmetric point. The number of copies of $V_{m+n-2s}$ equals (multiplicity of weight $m+n-2s$ ) minus (multiplicity of weight $m+n-2s+2$ ), which is 1 for $0\le s\le\min(m,n)$ and 0 beyond.

Exercise 7.5The Casimir is central

Show that $C=ef+fe+\frac12h^2$ commutes with $e$ , $f$ and $h$ , and conclude that it acts on $V_m$ by $\frac{m(m+2)}2$ .

Hint

Use $[X,YZ]=[X,Y]Z+Y[X,Z]$ . Then Schur's lemma and Example 7.5.

Exercise 7.6Representations of

\mathrm{SO}(3)

Show that $-I=\exp(i\pi h)$ in $\mathrm{SU}(2)$ , that it acts on $V_m$ by $(-1)^m$ , and deduce that the irreducible representations of $\mathrm{SO}(3)$ are exactly the $V_m$ with $m$ even, of odd dimension.

Hint

$\exp(i\pi h)=\mathrm{diag}(e^{i\pi},e^{-i\pi})$ . On $V_m$ , $h$ has eigenvalues $m-2k$ , and $e^{i\pi(m-2k)}=(-1)^m$ .

Exercise 7.7No nontrivial 1-dimensional representations

Show that every 1-dimensional representation of $\mathfrak{sl}(2)$ is zero. (Compare groups, whose 1-dimensional representations factor through the commutator quotient: here $\mathfrak{sl}(2)=[\mathfrak{sl}(2),\mathfrak{sl}(2)]$ .)

Hint

In dimension 1 all brackets vanish, so $h=[e,f]=0$ and $2e=[h,e]=0$ .

Exercise 7.8The adjoint representation is irreducible

Show that the adjoint representation of $\mathfrak{sl}(2)$ is irreducible, so $\mathfrak{sl}(2)$ has no ideals except $0$ and itself (it is simple).

Hint

A nonzero invariant subspace contains a weight vector ( $h$ acts diagonalizably), and from $e$ , $h$ or $f$ you reach the others by brackets.

InteractiveHall Ch. 6 §§1–7

SU(3): weights, roots and the Weyl group

For $\mathfrak{sl}(2)$ one diagonal element $h$ was enough to organize everything into a line of weights. For $\mathfrak{sl}(3)$ two commuting diagonal elements give weights in a plane, and the pictures become hexagons and triangles. The pattern scales to every semisimple Lie algebra, and here you can see it.

The idea

Cartan subalgebra and roots (Hall Ch. 6 §§1–2). In

\mathfrak{sl}(3,\mathbb C)

take

H_1=\mathrm{diag}(1,-1,0)

and

H_2=\mathrm{diag}(0,1,-1)

; they commute and span the diagonal subalgebra

\mathfrak h

. For

i\ne j

the matrix unit

E_{ij}

satisfies

[H,E_{ij}]=(h_i-h_j)E_{ij}

, so the six pairs

(h_i-h_j)

evaluated on

H_1,H_2

are the roots. They form a regular hexagon.

Weights. In a representation,

H_1

and

H_2

commute and act diagonalizably, so the space splits into simultaneous eigenspaces. The pair of eigenvalues

(m_1,m_2)

is a weight. Root vectors move weights by roots, exactly as

e

and

f

moved weights by

\pm2

for

\mathfrak{sl}(2)

.

Highest weight classification (Hall Ch. 6 §§3–4). Irreducible representations of

\mathfrak{sl}(3,\mathbb C)

, equivalently of

\mathrm{SU}(3)

, correspond to pairs of non-negative integers

(m_1,m_2)

, the highest weight. The dimension is

\frac12(m_1+1)(m_2+1)(m_1+m_2+2)

.

The Weyl group (Hall Ch. 6 §6). Permutation matrices in

\mathrm{SU}(3)

(up to sign) normalize the diagonal torus and act on weights by the symmetric group

S_3

, generated by the reflections

(m_1,m_2)\mapsto(-m_1,m_1+m_2)

and

(m_1,m_2)\mapsto(m_1+m_2,-m_2)

. Weight multiplicities are invariant under it, which is why the diagrams have six-fold or three-fold symmetry.

Explore: weight diagrams

Weights of the irreducible representation with highest weight

(m_1,m_2)

. A dot with

k

rings around it has multiplicity

k+1

. The faint hexagon marks the roots.

m₁ m₂

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Cartan subalgebra: A largest family of commuting, simultaneously diagonalizable elements; their joint eigenvalues label everything else.
Root: The pattern of eigenvalues by which the commuting diagonal elements act, through the bracket, on one of the off-diagonal elements of the algebra itself.
Weight: The list of joint eigenvalues of the commuting diagonal elements on one vector of a representation: a point in the plane, for this group.
Highest weight: The weight furthest out in a fixed preferred direction; it pins down an irreducible representation completely.
Weyl group: The group of symmetries of the weight picture coming from permuting coordinates; it permutes weights while keeping their multiplicities.
Multiplicity of a weight: How many independent vectors in a representation share the same list of joint eigenvalues.

Examples

Example 8.1 (the defining representation).

On $\mathbb C^3$ , $e_1,e_2,e_3$ have weights $(1,0)$ , $(-1,1)$ , $(0,-1)$ : a triangle, highest weight $(1,0)$ , dimension $\frac12\cdot2\cdot1\cdot3=3$ . Physicists call it the quark triplet.

Example 8.2 (the dual).

$(\mathbb C^3)^*$ has the negated weights $(-1,0)$ , $(1,-1)$ , $(0,1)$ : the inverted triangle, highest weight $(0,1)$ . In general the dual of $(m_1,m_2)$ is $(m_2,m_1)$ .

Example 8.3 (the adjoint, Hall Ch. 6 §5).

$\mathfrak{sl}(3)$ acting on itself has the six roots as weights, each once, and $(0,0)$ twice (spanned by $H_1,H_2$ ): a hexagon with a doubled centre, highest weight $(1,1)$ , dimension 8. This is the "eightfold way" of the meson octet.

Example 8.4 (symmetric powers).

$\mathrm{Sym}^k\mathbb C^3$ is irreducible with highest weight $(k,0)$ and dimension $\binom{k+2}2$ . For $k=3$ it is the 10-dimensional triangle of the baryon decuplet, every weight with multiplicity 1.

Example 8.5 (tensor products).

$\mathbb C^3\otimes(\mathbb C^3)^*\cong\mathfrak{sl}(3)\oplus\mathbb C$ , dimensions $9=8+1$ (traceless matrices plus scalars), and $\mathbb C^3\otimes\mathbb C^3=\mathrm{Sym}^2\oplus\Lambda^2\cong(2,0)\oplus(0,1)$ , dimensions $9=6+3$ .

Check your understanding

What is the dimension of the irreducible representation with highest weight $(2,1)$ ?

8That's $(1,1)$ .
15
10That's $(3,0)$ .

\frac12\cdot3\cdot2\cdot5=15

. Set

m_1=2

,

m_2=1

in the figure to see it.

Why is the zero weight of the adjoint representation doubled?

The zero weight space is the Cartan subalgebra itself, which is 2-dimensional.
Because the Weyl group has order 6.The Weyl group fixes the zero weight; it doesn't create multiplicity.

[H,X]=0

for all diagonal

H

exactly when

X

is diagonal, and the traceless diagonal matrices are 2-dimensional.

Exercises

Exercise 8.1The roots of

\mathfrak{sl}(3)

Show $[H,E_{ij}]=(h_i-h_j)E_{ij}$ for diagonal $H$ , and list the six roots as pairs (eigenvalue of $H_1$ , eigenvalue of $H_2$ ).

Hint

$HE_{ij}=h_iE_{ij}$ and $E_{ij}H=h_jE_{ij}$ . The roots are $\pm(2,-1)$ , $\pm(-1,2)$ , $\pm(1,1)$ .

Solution

$[H,E_{ij}]=(h_i-h_j)E_{ij}$ . For $H_1=\mathrm{diag}(1,-1,0)$ and $H_2=\mathrm{diag}(0,1,-1)$ : $E_{12}$ gives $(1-(-1),\,0-1)=(2,-1)$ ; $E_{23}$ gives $(-1-0,\,1-(-1))=(-1,2)$ ; $E_{13}$ gives $(1,1)$ ; and $E_{ji}$ gives the negatives.

Exercise 8.2Root vectors move weights by roots

Show that if $v$ has weight $\mu$ and $X\in\mathfrak g$ satisfies $[H,X]=\alpha(H)X$ , then $Xv$ has weight $\mu+\alpha$ (or is 0).

Hint

The same one-line computation as Exercise 7.1 for $\mathfrak{sl}(2)$ : $HXv=[H,X]v+XHv$ .

Solution

$H(Xv)=[H,X]v+X(Hv)=\alpha(H)Xv+\mu(H)Xv$ .

Exercise 8.3The adjoint representation

Verify Example 8.3: the weights of $\mathrm{ad}$ are the roots and 0, and $E_{13}$ is a highest weight vector of weight $(1,1)$ .

Hint

"Highest" means $E_{12}$ and $E_{23}$ kill it: $[E_{12},E_{13}]=[E_{23},E_{13}]=0$ .

Exercise 8.4Checking the dimension formula

Compute $\frac12(m_1+1)(m_2+1)(m_1+m_2+2)$ for $(1,0),(0,1),(1,1),(2,0),(3,0),(2,1),(2,2)$ , and check each against the count of weights with multiplicity in the figure.

Hint

$3,3,8,6,10,15,27$ . For $(2,2)$ the centre has multiplicity 3 and the inner hexagon multiplicity 2.

Exercise 8.5The Weyl group

Show the two reflections in the text generate a group of order 6 permuting the six roots, and that it is the action of $S_3$ permuting the diagonal entries $h_1,h_2,h_3$ .

Hint

A weight is determined by the eigenvalues $(h_1,h_2,h_3)$ with $m_1=h_1-h_2$ , $m_2=h_2-h_3$ . Swapping $h_1,h_2$ gives the first reflection.

Exercise 8.6Duals

Show that the weights of $V^*$ are the negatives of those of $V$ , and deduce that the dual of the irreducible with highest weight $(m_1,m_2)$ has highest weight $(m_2,m_1)$ .

Hint

The lowest weight of $V$ is the image of the highest under the longest Weyl group element, which sends $(m_1,m_2)$ to $(-m_2,-m_1)$ .

Exercise 8.7

3\otimes3

Using weights, show $\mathbb C^3\otimes\mathbb C^3\cong(2,0)\oplus(0,1)$ .

Hint

The nine weights are sums of two of $(1,0),(-1,1),(0,-1)$ . Remove the six weights of $(2,0)$ ; the three left are those of $(0,1)$ .

ReadingTu, Connections, Curvature, and Characteristic Classes, Ch. 5 §22 and Ch. 6 §§27–32

Lie groups in differential geometry

Lie groups are also the symmetries of geometric spaces, and their Lie algebras are where connection and curvature forms take their values. This module is the hand-off to the Connections & Curvature page: the Maurer–Cartan form, invariant metrics, homogeneous spaces and principal bundles.

The idea

The Maurer–Cartan form. On a matrix group,

\omega=g^{-1}dg

is a

\mathfrak g

-valued 1-form: at

g

it sends a tangent vector

v

to

g^{-1}v\in\mathfrak g

, inverting the construction of left-invariant fields. It satisfies the structure equation

d\omega+\omega\wedge\omega=0

(Exercise 9.1). Compare the curvature of a connection,

\Omega=dA+A\wedge A

on the Connections & Curvature page: the Maurer–Cartan form is a connection with zero curvature.

Invariant metrics. An inner product on

\mathfrak g

spreads by left translation to a left-invariant Riemannian metric. It is also right-invariant (bi-invariant) when the inner product is

\mathrm{Ad}

-invariant, and every compact group has such an inner product (average over the group, as in Maschke's theorem). For a bi-invariant metric the Levi-Civita connection is

\nabla_XY=\frac12[X,Y]

on left-invariant fields, the geodesics through

I

are the one-parameter subgroups

e^{tX}

, and the sectional curvature is

\frac14\lVert[X,Y]\rVert^2

for orthonormal

X,Y

: never negative.

Homogeneous spaces. If

G

acts transitively on a manifold

M

and

H

is the stabilizer of a point, then

M\cong G/H

. Spheres are

S^n=\mathrm{SO}(n+1)/\mathrm{SO}(n)

, projective spaces are quotients of unitary groups, and hyperbolic space is

\mathrm{SO}^+(n,1)/\mathrm{SO}(n)

. The geometry of

M

is then Lie algebra.

Principal bundles. A principal

G

-bundle

P\to M

is a family of copies of

G

parametrized by

M

, such as the frame bundle of an oriented surface (

G=\mathrm{SO}(2)

). A connection is a

\mathfrak g

-valued 1-form on

P

that restricts to the Maurer–Cartan form on each fibre; in a local trivialization it is a

\mathfrak g

-valued form

A

on

M

, which changes by

A\mapsto g^{-1}Ag+g^{-1}dg

under a change of frame (Exercise 9.5). Curvature, holonomy and Chern–Weil theory all live here: see Principal bundles and holonomy.

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Maurer–Cartan form: The rule that takes a velocity at any point of the group and translates it back to the identity, giving an infinitesimal motion there.
Bi-invariant metric: A way of measuring lengths on the group that is unchanged by multiplying on either side; compact groups always have one.
Homogeneous space: A space on which a group acts so that any point can be moved to any other; it is the group divided by the subgroup fixing one point.
Principal bundle: A space that looks locally like (piece of the base) × (the group), with the group acting on itself along each fibre; frames of a tangent space are the standard example.
Gauge transformation: A change of frame that varies from point to point, under which a connection form picks up an extra term from the derivative of the change.

Examples

Example 9.1 (the circle).

On $\mathrm U(1)$ , $g=e^{i\theta}$ gives $g^{-1}dg=i\,d\theta$ . So the Maurer–Cartan form is the familiar $d\theta$ (times $i$ ), and $d\omega=0=\omega\wedge\omega$ .

Example 9.2 (the affine group).

For $g=\begin{pmatrix}a&b\\0&1\end{pmatrix}$ , $g^{-1}dg=\begin{pmatrix}da/a&db/a\\0&0\end{pmatrix}$ . The left-invariant metric making $da/a$ and $db/a$ orthonormal is $\frac{da^2+db^2}{a^2}$ : the hyperbolic plane. (It is not right-invariant; this group is not compact.)

Example 9.3 (the round 3-sphere).

On $\mathrm{SU}(2)\cong S^3$ , the bi-invariant metric from $\langle X,Y\rangle=-2\,\mathrm{tr}(XY)$ is (a multiple of) the round metric, and its geodesics through $I$ are great circles $e^{tX}$ , as in the right panel of module 6's figure.

Example 9.4 (the sphere as a homogeneous space).

$\mathrm{SO}(3)$ acts transitively on $S^2$ , and the stabilizer of the north pole is the rotations about the $z$ -axis, so $S^2\cong\mathrm{SO}(3)/\mathrm{SO}(2)$ , with $\dim=3-1$ . Choosing a unit tangent vector as well identifies $\mathrm{SO}(3)$ with the unit tangent bundle of $S^2$ .

Example 9.5 (frames of a surface).

Orthonormal frames on an oriented Riemannian surface form a principal $\mathrm{SO}(2)$ -bundle. The Levi-Civita connection is a $\mathfrak{so}(2)\cong\mathbb R$ -valued form $\omega_{12}$ , and its curvature $d\omega_{12}$ is $K$ times the area form: the setting of Gauss–Bonnet on the Connections & Curvature page.

Check your understanding

For a bi-invariant metric on a compact Lie group, the geodesics through the identity are:

the one-parameter subgroups $t\mapsto e^{tX}$ .
straight lines in the space of matrices.Those leave the group.
only the closed one-parameter subgroups.Dense windings on a torus are geodesics too.

\nabla_XX=\frac12[X,X]=0

for a left-invariant

X

, so its integral curves

ge^{tX}

are geodesics (Exercise 9.3).

The Maurer–Cartan form satisfies $d\omega+\omega\wedge\omega=0$ . In the language of connections this says:

it is a flat connection: its curvature vanishes.
it is the Levi-Civita connection.It is a connection on the trivial bundle, not on the tangent bundle with a metric.

The curvature of a connection form

A

is

dA+A\wedge A

.

Exercises

Exercise 9.1The structure equation

Show that $\omega=g^{-1}dg$ satisfies $d\omega+\omega\wedge\omega=0$ .

Hint

Differentiate $g^{-1}g=I$ : $d(g^{-1})=-g^{-1}\,dg\,g^{-1}$ .

Solution

$d\omega=d(g^{-1})\wedge dg=-g^{-1}dg\,g^{-1}\wedge dg=-(g^{-1}dg)\wedge(g^{-1}dg)=-\omega\wedge\omega$ , using $d(dg)=0$ .

Exercise 9.2Bi-invariant metrics exist on compact groups

Show that a left-invariant metric is right-invariant iff its inner product on $\mathfrak g$ is $\mathrm{Ad}$ -invariant, and that every compact group has one.

Hint

Right translation by $h$ acts at the identity, after moving back by left translation, as $\mathrm{Ad}_{h^{-1}}$ . For existence, average any inner product over $G$ with Haar measure, exactly as in the unitary trick on the representation theory page.

Exercise 9.3One-parameter subgroups are geodesics

Given that the Levi-Civita connection of a bi-invariant metric is $\nabla_XY=\frac12[X,Y]$ on left-invariant fields, show that $t\mapsto ge^{tX}$ is a geodesic.

Hint

The curve is an integral curve of $X^L$ , so its acceleration is $\nabla_{X^L}X^L=\frac12[X^L,X^L]=0$ .

Exercise 9.4Spheres as homogeneous spaces

Show that $\mathrm{SO}(n+1)$ acts transitively on $S^n$ with stabilizer $\mathrm{SO}(n)$ , and deduce $\dim\mathrm{SO}(n+1)-\dim\mathrm{SO}(n)=n$ . Check against Example 4.5.

Hint

Any unit vector is the first column of some rotation (complete it to an oriented orthonormal basis). $\frac{(n+1)n}2-\frac{n(n-1)}2=n$ .

Exercise 9.5The gauge transformation rule

A connection on a trivial bundle is a matrix-valued 1-form $A$ , and parallel sections $s$ satisfy $ds+As=0$ . Under a change of frame $s=g\tilde s$ , show the new connection form is $\tilde A=g^{-1}Ag+g^{-1}dg$ , and that curvature transforms by $\tilde\Omega=g^{-1}\Omega g$ .

Hint

Substitute $s=g\tilde s$ into $ds+As=0$ and multiply by $g^{-1}$ . For curvature, use Exercise 9.1. Compare "Change of frame and Bianchi" on the Connections & Curvature page.

Solution

$d(g\tilde s)+Ag\tilde s=g\,d\tilde s+(dg+Ag)\tilde s=0$ , so $d\tilde s+(g^{-1}dg+g^{-1}Ag)\tilde s=0$ . The curvature computation is a direct expansion of $d\tilde A+\tilde A\wedge\tilde A$ in which the terms involving $dg$ cancel because $g^{-1}dg$ satisfies the structure equation. Note the special case $A=0$ : the flat connection $g^{-1}dg$ is the Maurer–Cartan form.

Exercise 9.6The hyperbolic plane as a group

In Example 9.2, check that left translations $(a,b)\mapsto(a'a,a'b+b')$ are isometries of $\frac{da^2+db^2}{a^2}$ , and identify them with the maps $z\mapsto a'z+b'$ of the upper half-plane (with $z=b+ia$ ).

Hint

Under the map, $da\mapsto a'\,da$ , $db\mapsto a'\,db$ , $a\mapsto a'a$ . The affine group acts simply transitively on the hyperbolic plane, so as a manifold the group is the hyperbolic plane.

The idea

Explore: which groups contain this matrix?

Key terms in plain words

Examples

Check your understanding

Exercises

The idea

Explore: one-parameter subgroups of GL(2,ℝ)\mathrm{GL}(2,\mathbb R)

Key terms in plain words

Examples

Check your understanding

Exercises

The idea

Watch

Worked example

Examples

Check your understanding

Practice

The idea

Explore: invariant vector fields on the group of x↦ax+bx\mapsto ax+b

Key terms in plain words

Examples

Check your understanding

Exercises

The idea

Explore: how good is each truncation?

Key terms in plain words

Examples

Check your understanding

Exercises

The idea

Explore: turning twice

Key terms in plain words

Examples

Check your understanding

Exercises

The idea

Explore: weight diagrams and Clebsch–Gordan

Key terms in plain words

Examples

Check your understanding

Exercises

The idea

Explore: weight diagrams

Key terms in plain words

Examples

Check your understanding

Exercises

The idea

Key terms in plain words

Examples

Check your understanding

Exercises

Explore: one-parameter subgroups of $\mathrm{GL}(2,\mathbb R)$

Explore: invariant vector fields on the group of $x\mapsto ax+b$